# Expectation of F-Distribution

## Theorem

Let $n, m$ be strictly positive integers.

Let $X \sim F_{n, m}$ where $F_{n, m}$ is the F-distribution with $\tuple {n, m}$ degrees of freedom.

Then the expectation of $X$ is given by:

$\expect X = \dfrac m {m - 2}$

for $m > 2$, and does not exist otherwise.

## Proof

Let $Y$ and $Z$ be independent random variables.

Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.

$\dfrac {Y / n} {Z / m} \sim F_{n, m}$

So we aim to compute:

$\expect {\dfrac {Y / n} {Z / m} }$

Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.

Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$.

From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$:

$\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$

We therefore have:

 $\ds \expect {\dfrac {Y / n} {Z / m} }$ $=$ $\ds \int_0^\infty \int_0^\infty \frac {y / n} {z / m} \map {f_{Y, Z} } {y, z} \rd y \rd z$ $\ds$ $=$ $\ds \frac m n \int_0^\infty \int_0^\infty \frac y z \map {f_Y} y \map {f_Z} z \rd y \rd z$ $\ds$ $=$ $\ds \frac m n \paren {\int_0^\infty \frac {\map {f_Z} z} z \rd z} \paren {\int_0^\infty y \map {f_Y} y \rd y}$ rewriting $\ds$ $=$ $\ds \frac m n \paren {\frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z} \paren {\frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y}$ Definition of Chi-Squared Distribution

Note that the integral converges:

$\displaystyle \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$
$\dfrac m 2 - 2 > -1$

That is:

$m > 2$

With that, we have:

 $\ds \frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$ $=$ $\ds \frac 1 {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty \paren {2 u}^{m / 2 - 2} e^{-u} \rd u$ substituting $z = 2 u$ $\ds$ $=$ $\ds \frac {2^{m / 2 - 2} } {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty u^{m / 2 - 2} e^{-u} \rd u$ $\ds$ $=$ $\ds \frac 1 2 \frac {\map \Gamma {\frac m 2 - 1} } {\map \Gamma {\frac m 2} }$ Definition of Gamma Function $\ds$ $=$ $\ds \frac {\map \Gamma {\frac m 2 - 1} } {\paren {\frac m 2 - 1} \map \Gamma {\frac m 2 - 1} }$ Gamma Difference Equation $\ds$ $=$ $\ds \frac 1 {m - 2}$

The integral:

$\displaystyle \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$

converges if and only if $n > -2$.

This is ensured by the fact that $n \in \N$.

With that, we have:

 $\ds \frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$ $=$ $\ds \frac 2 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty \paren {2 v}^{n / 2} e^{-v} \rd v$ substituting $y = 2 v$ $\ds$ $=$ $\ds \frac {2^{n / 2} } {2^{n / 2 - 1} \map \Gamma {\frac n 2} } \int_0^\infty v^{n / 2} e^{-v} \rd v$ $\ds$ $=$ $\ds 2 \times \frac {\map \Gamma {\frac n 2 + 1} } {\map \Gamma {\frac n 2} }$ $\ds$ $=$ $\ds 2 \times \frac {\frac n 2 \map \Gamma {\frac n 2} } {\map \Gamma {\frac n 2} }$ Gamma Difference Equation $\ds$ $=$ $\ds n$

So:

 $\ds \expect {\dfrac {Y / n} {Z / m} }$ $=$ $\ds \frac {n m} {n \paren {m - 2} }$ $\ds$ $=$ $\ds \frac m {m - 2}$

$\blacksquare$